Question: Solve for $X$. $X+\left[\begin{array}{rr} 20 & 10 \\ -12 & 3 \end{array}\right]=\left[\begin{array}{rr} 23 & 16 \\ 24 & 11\end{array}\right] $ $X=$
The Strategy First, we can represent the matrices of the equation with letters, which will make the equation easier to handle. Then we can solve the equation for $X$ and obtain an expression with the letters we defined. Finally, we can substitute back the actual matrices into the resulting expression and simplify it. Solving the equation for $X$ We are given the following equation. $X+\left[\begin{array}{rr} 20 & 10 \\ -12 & 3 \end{array}\right]=\left[\begin{array}{rr} 23 & 16 \\ 24 & 11\end{array}\right] $ Let's represent the above matrices as follows. $A=\left[\begin{array}{rr} 20 & 10 \\ -12 & 3 \end{array}\right] ~~~~~~~~~ B = \left[\begin{array}{rr} 23 & 16 \\ 24 & 11\end{array}\right] $ Then we can rewrite the equation as follows. $X+A=B$ Now it's simple to solve the equation for $X$. $\begin{aligned}X+A&=B\\\\ X&=B-A \end{aligned}$ Finding $X$ We found that $X=B-A$. Now we can substitute the actual matrices back into the expression and simplify. $\begin{aligned}X&=B-A \\\\&=\left[\begin{array}{rr} 23 & 16 \\ 24 & 11\end{array}\right] -\left[\begin{array}{rr} 20 & 10 \\ -12 & 3 \end{array}\right] \\\\\\&=\left[\begin{array}{rr}(23-20) & (16-10) \\ (24+12) & (11-3) \end{array}\right] \\\\\\&=\left[\begin{array}{rr}3 & 6 \\ 36 & 8\end{array}\right]\end{aligned}$ Summary $X=\left[\begin{array}{rr}3 & 6 \\ 36 & 8\end{array}\right]$